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kolezko [41]
3 years ago
15

Divide 6,580 ÷ 31 A.212 r 2 B.212 r8 C.214 r2 D.214 r 8

Mathematics
1 answer:
Sloan [31]3 years ago
8 0
Your answer to the question is C.214 R2 hope you get it right
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G(x)=f(x)+k,solve for k
vesna_86 [32]

Step-by-step explanation:

Top answer · 0 votes

f(g(x))=362−x−2k f(g(x))=x→362−x−2k=x→x2−2x(1−k)−4(k−9)=0 Solution for x are equal if and only if discriminant Δ=0 4(1−k)2+16(k−9)=0 ...  More

3 0
3 years ago
Elvaluate the expression.<br>9!/3!<br>a) 3<br>b) 6<br>c) 60,480<br>d) 362,874​
HACTEHA [7]

Answer:

c) 60,480

Step-by-step explanation:

9×8×7×6×5×4×3×2×1

÷

3×2×1

Cancel same numbers in the numerator and denominator.

So you're going to have= 9×8×7×6×5×4 divided by 1

Which gives you, 60,480.

Hope this helped:) Mark me brainiest please.

4 0
3 years ago
The heights of a random sample of 50 college students showed a mean of 174.5 centimeters and a standard deviation of 6.9 centime
Minchanka [31]

Answer:

Step-by-step explanation:

Hello!

For me, the first step to any statistics exercise is to determine what is the variable of interest and it's distribution.

In this example the variable is:

X: height of a college student. (cm)

There is no information about the variable distribution. To estimate the population mean you need a variable with at least a normal distribution since the mean is a parameter of it.

The option you have is to apply the Central Limit Theorem.

The central limit theorem states that if you have a population with probability function f(X;μ,δ²) from which a random sample of size n is selected. Then the distribution of the sample mean tends to the normal distribution with mean μ and variance δ²/n when the sample size tends to infinity.

As a rule, a sample of size greater than or equal to 30 is considered sufficient to apply the theorem and use the approximation.

The sample size in this exercise is n=50 so we can apply the theorem and approximate the distribution of the sample mean to normal:

X[bar]~~N(μ;σ2/n)

Thanks to this approximation you can use an approximation of the standard normal to calculate the confidence interval:

98% CI

1 - α: 0.98

⇒α: 0.02

α/2: 0.01

Z_{1-\alpha /2}= Z_{1-0.01}= Z_{0.99} =2.334

X[bar] ± Z_{1-\alpha /2} * \frac{S}{\sqrt{n} }

174.5 ± 2.334* \frac{6.9}{\sqrt{50} }

[172.22; 176.78]

With a confidence level of 98%, you'd expect that the true average height of college students will be contained in the interval [172.22; 176.78].

I hope it helps!

4 0
3 years ago
Helena lost her marbles. But then she found them and put them in 4 bags with mm marbles in each bag. She had 3 marbles left over
77julia77 [94]

Answer:

4m +3 marbles does Helena have in all.

Step-by-step explanation:

Here, m represents the marble in each bag.

She put them in 4 bags with m marbles in each bag.

In each bag she have m marbles

then, in 4 bag she have 4m marbles

⇒total number of marbles= 4m

Since, she had 3 marbles left over.

⇒She have total number of marbles = 4m + 3

Therefore, 4m +3 marbles does Helena have in all.


8 0
3 years ago
Read 2 more answers
Complete each ratio table
Nesterboy [21]
?????????? is there supposed to be an attachment because its not there.
5 0
3 years ago
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